Question: $\sum\limits_{n=1}^{\infty} \dfrac{1}{\sqrt{n+3}}$ When applying the integral test, we get a limit that determines whether the series converges or diverges. What is this limit? Choose 1 answer: Choose 1 answer: (Choice A) A $\lim_{b\to\infty} \sqrt{b+3} $ (Choice B) B $\lim_{b\to\infty} \left[ 2\sqrt{b+3} - 4\right]$ (Choice C) C $\lim_{b\to\infty} \left[\ln \sqrt{b+3} - \ln{2}\right]$ (Choice D) D $ \lim_{b\to\infty} \dfrac{1}{\sqrt{b+3}}$
Solution: $\dfrac{1}{\sqrt{n+3}}$ satisfies the conditions for the integral test. This means that $\sum\limits_{n=1}^{\infty} \dfrac{1}{\sqrt{n+3}}$ converges/diverges together with $\int_1^{\infty} \dfrac{1}{\sqrt{x+3}} \,dx$. $\int_1^{\infty} \dfrac{1}{\sqrt{x+3}} \,dx=\lim_{b\to\infty} \left[ 2\sqrt{b+3} - 4\right]$ In conclusion, the limit that determines whether the series converges or diverges is $\lim_{b\to\infty} \left[ 2\sqrt{b+3} - 4\right]$.